Problem TPDE

Engineering Mathematics Material 2010

Prepared by C.Ganesan, M.Sc., M.Phil., (Ph: 9841168917)

SUBJECT NAME : Transforms and Partial Differential Equations

SUBJECT CODE : MA 2211

MATERIAL NAME : Problem Material

MATERIAL CODE : JM08AM3006

Name of the Student: Branch: Unit – I (Fourier Series)

1) Develop a Fourier series for the function ( ) ( )f x x xππππ= −= −= −= − in the interval (0,2 )ππππ .

2) Find the Fourier series for the function 2( ) ( )f x xππππ= −= −= −= − in the interval (0,2 )ππππ and

deduce sum of series as 2 2 2

1 1 11 ...

2 3 4+ + + ++ + + ++ + + ++ + + + and also find 4 4 4

1 1 11 ...

2 3 4+ + + ++ + + ++ + + ++ + + +

3) Find the Fourier series of periodic 2π for in (0, )

( )2 in ( ,2 )

xf x

x

πππππ π ππ π ππ π ππ π π

==== −−−−

and deduce that

2

2 2 2

1 1 1...

1 3 5 8ππππ+ + + =+ + + =+ + + =+ + + = .

4) Obtain the Fourier expansion of periodicity 2π for ( )f x x==== when xπ ππ ππ ππ π− < <− < <− < <− < < .

Deduce that

2

2 2

1 11 ...

3 5 8ππππ+ + + =+ + + =+ + + =+ + + = and also find the value of 4 4

1 11 ...

3 5+ + ++ + ++ + ++ + +

5) Find the Fourier series of 2( )f x x==== in the interval ( , )π ππ ππ ππ π−−−− . Hence deduce that

2 2 2

1 1 1...

1 2 3+ + ++ + ++ + ++ + + also find the value of 4 4 4

1 1 1...

1 2 3+ + ++ + ++ + ++ + +

6) Find the Fourier series for the function2( )f x x x= += += += + in the interval ( , )π ππ ππ ππ π−−−− and also

deduce the value of

2

2

16n

ππππ====∑∑∑∑ .

7) Show that when 0 x ππππ< << << << < , sin 2 sin4 sin6

...2 1 2 3

x x xx

ππππππππ − = + + +− = + + +− = + + +− = + + +

8) Develop a sine series of the function 0 / 2

( ) / 2

x xf x

x x

πππππ π ππ π ππ π ππ π π

< << << << <==== − < <− < <− < <− < <

9) Find the Fourier series of ( ) cosf x x x==== in the interval ( , )π ππ ππ ππ π−−−− .

10) Find the Fourier series of ( ) cosf x x==== in the interval [-π, π].

11) Obtain the Fourier series of ( )f x period 2 � and defined as follows

0( )

0 2

x xf x

x

− < <− < <− < <− < <==== < << << << <

� ��

� �� and hence deduce the following



i) 1 1 1

1 ...3 5 7 4

ππππ− + − + ∞ =− + − + ∞ =− + − + ∞ =− + − + ∞ =

ii) 2

2 2 2

1 1 1...

1 3 5 8ππππ+ + + ∞ =+ + + ∞ =+ + + ∞ =+ + + ∞ =

12) Find the Fourier series of 2( )f x x==== in the interval ( , )−−−−� �� . Hence deduce the sum of

series 2 2 2

1 1 1...

1 2 3+ + + ∞+ + + ∞+ + + ∞+ + + ∞ .

13) Find the Fourier series of ( ) (2 )f x x x= −= −= −= − in 0 2x< << << << < . Deduce the sum of the

series 2 2 2 2

1 1 1 1...

1 2 3 4+ + + + ∞+ + + + ∞+ + + + ∞+ + + + ∞ .

14) Find the complex form of the Fourier series of the function ( ) xf x e==== when

xπ ππ ππ ππ π− < <− < <− < <− < < and ( 2 ) ( )f x f xππππ+ =+ =+ =+ = .

15) Find the complex form of the Fourier series of the periodic function ( ) sinf x x==== when

0 x ππππ< << << << < .

16) Find the Fourier Series up to three harmonic for y = f(x)in (0,2π) for the following data

x 0 π/3 2π/3 π 4π/3 5π/3 2π

y 1.0 1.4 1.9 1.7 1.5 1.2 1.0

17) Find the Fourier Series up to two harmonic for y = f(x)in (0,2π) for the following data

x 0° 60° 120° 180° 240° 300° 360°

y 40.0 31.0 -13.7 20.0 3.7 -21.0 40.0

18) Find the Fourier Series up to two harmonic for y = f(x)in (0,2π) for the following data

x 0 1 2 3 4 5

y 9 18 24 28 26 20

19) The values of x and the corresponding values of f(x) over a period T are given below

Show that ( ) 0.75 0.37cos 1.004sinf x θ θθ θθ θθ θ= + += + += + += + + where2 xTππππθθθθ ==== .

x 0 T/6 T/3 T/2 2T/3 5T/6 T

f(x) 1.98 1.30 1.05 1.30 -0.88 -0.25 1.98



Unit – II (Fourier Transforms) Part – I

1) Show that the Fourier transform of

2 2, ( )

0 ,

a x x af x

x a

− <− <− <− <==== >>>>

is

3

2 sin cos2

as as assππππ

−−−− .

Hence deduce that 30

sin cos4

t t tdt

tππππ∞∞∞∞ −−−− ====∫∫∫∫ . Using

Parseval’s identity show that

2

30

sin cos15

t t tdt

tππππ∞∞∞∞ −−−− ====

∫∫∫∫ .

2) Find the Fourier transform of

21 , 1( )

0 , 1

x xf x

x

− <− <− <− <==== >>>>

. Hence prove that

30

sin cos 3cos

2 16s s s s

dss

ππππ∞∞∞∞ −−−− ====∫∫∫∫ .

3) Find the Fourier transform of ( )f x if1 ,

( )0 , 0

x af x

x a

<<<<==== > >> >> >> >

. Hence deduce that

i)

0

sin2

tdt

tππππ∞∞∞∞

====∫∫∫∫ ii)

2

0

sin2

tdt


==== ∫∫∫∫

4) Find the Fourier transform of ( )f x if1 , 1

( )0 , 1

x xf x

x

− <− <− <− <==== >>>>

. Hence deduce that

2

0

sin2

tdt


==== ∫∫∫∫

and

4

0

sin3

tdt


==== ∫∫∫∫ .

Note: The same problem they may ask ,

( )0 ,

a x x af x

x a

− <− <− <− <==== >>>>

with same

deduction.

Part – II

5) Find the Fourier cosine and sine transform of the function ( ) , 0axf x e a−−−−= >= >= >= > .

6) Evaluate

(((( ))))22 20

, 0dx

if aa x

∞∞∞∞

>>>>++++

∫∫∫∫ using Parseval’s identity.



7) Evaluate

(((( ))))2

22 20

, 0x

dx aa x

∞∞∞∞

>>>>++++

∫∫∫∫ using Parseval’s identity.

8) Evaluate (((( )))) (((( ))))2 2 2 20

, , 0dx

if a bx a x b

∞∞∞∞

>>>>+ ++ ++ ++ +∫∫∫∫ using Parseval’s identity.

9) Evaluate (((( )))) (((( ))))2

2 2 2 20

, , 0x

dx if a bx a x b

∞∞∞∞

>>>>+ ++ ++ ++ +∫∫∫∫ using Parseval’s identity.

10) Evaluate (((( )))) (((( ))))2 20 1 4

dx

x x

∞∞∞∞

+ ++ ++ ++ +∫∫∫∫ using transforms method.

Part – III (Special Problems)

11) Find the Fourier cosine and sine transform of ( ) axf x xe−−−−==== .

12) Find the Fourier cosine transform of ( )axe

f xx

−−−−

==== .

13) Find the Fourier sine transform of ( )axe

f xx

−−−−

==== .

14) Find the Fourier cosine transform of2

( ) xf x e−−−−==== .

15) Find the Fourier cosine transform of 2 2

, 0a xe a−−−− >>>> and hence deduce that sine

transform of 2 2a xxe−−−−

.

16) Show that the Fourier sine transform of

2

2x

xe−−−−

is self reciprocal.

17) Find the Fourier sine and cosine transform of xe−−−−

also find the Fourier sine transform of

21xx++++

and Fourier cosine transform of 2

11 x++++

.

18) State and Prove Convolution theorem in Fourier transform.

Unit – III (Partial Differential Equations)

• Higher Order Homogeneous PDE

1) Solve (((( ))))2 26 5 sinhxD DD D z e y xy′ ′′ ′′ ′′ ′− + = +− + = +− + = +− + = + .

Ans.:

3 4

1 2

1( ) ( 5 )

8 24 6 4x y x yx x y x

z f y x f y x e e+ −+ −+ −+ −= + + + − − + += + + + − − + += + + + − − + += + + + − − + + .

2) Solve (((( ))))3 2 24 4 6sin(3 6 )D D D DD z x y′ ′′ ′′ ′′ ′− + = +− + = +− + = +− + = + .

Ans.: 1 2 3

2( ) ( 2 ) ( 2 ) cos(3 6 )

81z f y f y x xf y x x y= + + + + + += + + + + + += + + + + + += + + + + + + .



3) Solve (((( ))))2 sin sin 2D DD z x y′′′′− =− =− =− = .

Ans.: 1 2

1 1( ) ( ) cos( 2 ) cos( 2 )

6 2z f y f y x x y x y= + + − − − += + + − − − += + + − − − += + + − − − + .

4) Solve

3 3 33

3 2 37 6 sin( 2 ) x yz z zz x y e

x x y y++++ ∂ ∂ ∂∂ ∂ ∂∂ ∂ ∂∂ ∂ ∂

− − = + +− − = + +− − = + +− − = + + ∂ ∂ ∂ ∂∂ ∂ ∂ ∂∂ ∂ ∂ ∂∂ ∂ ∂ ∂ .

Ans.: 3

1 2 3

1( ) ( 2 ) ( 3 ) cos( 2 )

75 20x yx

z f y x f y x f y x x y e ++++= − + − + + − + += − + − + + − + += − + − + + − + += − + − + + − + + .

5) Solve (((( ))))2 26 cosD DD D z y x′ ′′ ′′ ′′ ′+ − =+ − =+ − =+ − = .

Ans.: 1 2( 2 ) ( 3 ) cos sinz f y x f y x y x x= + + − − += + + − − += + + − − += + + − − + .

6) Solve (((( ))))2 22 2 2 sin( 2 )D DD D D D z x y′ ′ ′′ ′ ′′ ′ ′′ ′ ′+ + − − = ++ + − − = ++ + − − = ++ + − − = + .

7) Solve (((( ))))2 2 23 3 7x yD D D D z e xy++++′ ′′ ′′ ′′ ′− − + = + +− − + = + +− − + = + +− − + = + + .

8) Solve (((( ))))2 22 3 3 2 cosh(2 )D DD D D D z x y′ ′ ′′ ′ ′′ ′ ′′ ′ ′− + − + + = +− + − + + = +− + − + + = +− + − + + = + .

9) Solve (((( ))))2 22 2 2 cos 2 cosD DD D D D z x y′ ′ ′′ ′ ′′ ′ ′′ ′ ′− − + + =− − + + =− − + + =− − + + =

• Standard Types

1) Solve z px qy pq= + += + += + += + + . Ans.: 0z xy+ =+ =+ =+ =

2) Solve2 2z px qy p q= + + −= + + −= + + −= + + − . Ans.:

2 24z y x= −= −= −= −

3) Solve2 21z px qy p q= + + + += + + + += + + + += + + + + . Ans.:

2 2 2 1x y z+ + =+ + =+ + =+ + = .

4) Solve2 21z px qy c p q= + + + += + + + += + + + += + + + + . Ans.:

2 2 2 2x y z c+ + =+ + =+ + =+ + = .

5) Solvez x y

pqpq q p

= + += + += + += + + . Ans.:

Hint: Multiply pq on both side, we get (((( ))))3/2z px qy pq= + += + += + += + +

6) Solve (((( ))))2 2 21p y x qx+ =+ =+ =+ = .

Hint: Rearrange the above equation, we get (((( ))))2 2

2

1 (says)

p x qk

x y

++++= == == == = .

This is of the form 1 2( , ) ( , )f x p f y q==== .

7) Solve (((( ))))2 29 4p z q+ =+ =+ =+ = . Ans.: (((( )))) (((( ))))3 22z a x ay b+ = + ++ = + ++ = + ++ = + + .

Hint: This is of the form ( , , ) 0f z p q ==== .

8) Solve2 2 21z p q= + += + += + += + + . Ans.: (((( ))))1

2

1cosh

1z x ay b

a−−−− = + += + += + += + +

++++.




9) Solve (((( ))))1p q qz+ =+ =+ =+ = . Ans.: (((( ))))log 1x ay c az + = −+ = −+ = −+ = − .


10) Solve (((( )))) (((( )))) (((( ))))x y z p y z x q z x y− + − = −− + − = −− + − = −− + − = − . Ans.: (((( )))), 0x y z xyzφφφφ + + =+ + =+ + =+ + =

11) Solve (((( )))) (((( )))) (((( ))))2 2 2 2 2 2x y z p y z x q z x y− + − = −− + − = −− + − = −− + − = − . Ans.: (((( ))))2 2 2, 0x y z xyzφφφφ + + =+ + =+ + =+ + =

12) Solve (((( )))) (((( )))) (((( ))))m n n mz y p x z q y x− + − = −− + − = −− + − = −− + − = −� �� .

Ans.: (((( ))))2 2 2, m n 0x y z x y zφφφφ + + + + =+ + + + =+ + + + =+ + + + =��

13) Solve (((( )))) (((( )))) (((( ))))3 4 4 2 2 3z y p x z q y x− + − = −− + − = −− + − = −− + − = − .

Ans.: (((( ))))2 2 2,2 3 4 0x y z x y zφφφφ + + + + =+ + + + =+ + + + =+ + + + =

14) Solve (((( )))) (((( )))) (((( ))))2 2 2x yz p y zx q z xy− + − = −− + − = −− + − = −− + − = − .

Ans.: , 0x y

xy yz zxy z

φφφφ −−−− + + =+ + =+ + =+ + = −−−−

Unit – IV (Application of PDE) • One dimensional Wave equation.

1) Problems on Zero initial velocity (Pre work: 1).

For the following ( )f x

a) ( ) sinx

f x aππππ====��

(another name Sinusoid of length a)

b) 30( ) sin

xf x y

ππππ====��

c) ( ) ( )f x x xλλλλ= −= −= −= −��

d) The midpoint of the string is displaced to a small height ‘h’ or ‘b’ is given.

e) Any problems with string of length 2ℓ.

2) Problems on Non – Zero initial velocity (Pre work: 2).

For the following ( )f x

f) ( ) sinx

f x aππππ====��

(another name Sinusoid of length a)

g) 30( ) sin

xf x v

ππππ====��

h) ( ) ( )f x x xλλλλ= −= −= −= −��

i) ,0 / 2

( ), / 2

cx xv

c x x

≤ ≤≤ ≤≤ ≤≤ ≤==== − ≤ ≤− ≤ ≤− ≤ ≤− ≤ ≤

��

� � ��

j) Any problems with string of length 2ℓ.



• One dimensional Heat flow equation.

3) Problems on zero boundary conditions. (Pre work: 3)

4) Problems on One end zero and another end non – zero boundary condition and

reduced to 0°C. (Pre work: 4)

5) Problems on both ends non – zero boundary condition and reduced to 0°C.

6) Problems on both ends non – zero boundary condition and reduced to non – zero

temperature.

Unit – V (Z – Transform) • Problems on Z – transform

1) Find the Z – transform of the function cos , sin , cos , sin ,n nn n a n a nθ θ θ θθ θ θ θθ θ θ θθ θ θ θ

cos / 2, sin / 2, cos / 2, sin / 2.n nn n a n a nπ π π ππ π π ππ π π ππ π π π

2) State and Prove final value theorem in Z – transform.

3) State and Prove Initial value theorem in Z – transform.

4) Find the Z – transform of the following functions (i) {{{{ }}}}na (ii) {{{{ }}}}nna (iii) 1n

(iv)

1!n

(v) !

nan

(vi) 1

1n ++++

(vii) {{{{ }}}}n (viii) 1n

(ix) (((( )))) (((( ))))1 2n n+ ++ ++ ++ + (x) (((( ))))1

1n n ++++

5) Find the Z – transform of the function (((( )))) (((( ))))2 3

( )1 2n

f nn n

++++====+ ++ ++ ++ +

.

• Problems on Inverse Z – transform using Partial fraction

method

1) Find the inverse Z – transform of (((( )))) (((( )))) (((( ))))2 3

2 1 5z z

z z z++++

+ − ++ − ++ − ++ − + using partial fraction

method.

2) Find (((( ))))

(((( )))) (((( ))))

2

12

2

1 1

z z zZ

z z−−−− − +− +− +− + + −+ −+ −+ −

by using partial fraction method.

3) Find the inverse Z – transform of(((( )))) (((( ))))

2

22 4

z

z z+ ++ ++ ++ + using partial fraction method.

4) Find 1

2 7 10z

Zz z

−−−− + ++ ++ ++ +

by using partial fraction method.

5) Find the inverse Z – transform of

2

3 2

33 4

z zz z

++++− +− +− +− +

using partial fraction method.



6) Find (((( ))))

(((( ))))1

3

1

1

z zZ

z−−−− ++++

−−−− by using reduces method.

• Problems on Convolution Theorem

1) State Convolution theorem and use it to evaluate (((( ))))2

1

( )z

Zz a z b

−−−− − −− −− −− −

.

2) Find

21

2( )z

Zz a

−−−− ++++

using convolution theorem.

3) Find

21 8

(2 1)(4 1)z

Zz z

−−−− − +− +− +− +

using convolution theorem.

4) Using Convolution theorem, find (((( )))) (((( ))))2

1

1 3z

Zz z

−−−− − −− −− −− −

.

5) Find the inverse Z – transform of

2z

z a −−−−

, using Convolution theorem.

6) Use Convolution theorem to find the inverse Z – transform (((( )))) (((( ))))1 1

12

3 4z z− −− −− −− −− −− −− −− −.

• Solving Difference Equation

1) Using Z – transform, solve 2 16 9 2kk k ky y y+ ++ ++ ++ ++ + =+ + =+ + =+ + = given 0 1 0y y= == == == = .

2) Solve 2 13 2 0n n nu u u+ ++ ++ ++ ++ + =+ + =+ + =+ + = given 0 11, 2u u= == == == = , using Z – transform.

3) Solve ( 3) 3 ( 1) 2 ( ) 0y n y n y n+ − + + =+ − + + =+ − + + =+ − + + = given that (0) 4, (1) 0y y= == == == = and

(2) 8.y ====

4) Solve the system 2 15 6 ,n n n ny y y u+ ++ ++ ++ +− + =− + =− + =− + = with 0 10, 1y y= == == == = and 1nu ==== for

0,1,2, ...n ==== by Z – transform method.

Hint: Substitute 1nu ==== in given equation, we get 2 15 6 1n n ny y y+ ++ ++ ++ +− + =− + =− + =− + = and do as

usual method.

5) Using Z – transform solve ( ) 3 ( 1) 4 ( 2) 0,y n y n y n+ − − − =+ − − − =+ − − − =+ − − − = 2n ≥≥≥≥ given that

(0) 3y ==== and (1) 2.y = −= −= −= −

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Problem TPDE

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